Factor the following expression: $2$ $x^2+$ $9$ $x$ $-5$
Solution: This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(-5)} &=& -10 \\ {a} + {b} &=& & & {9} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-10$ and add them together. Remember, since $-10$ is negative, one of the factors must be negative. The factors that add up to ${9}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${10}$ $ \begin{eqnarray} {ab} &=& ({-1})({10}) &=& -10 \\ {a} + {b} &=& {-1} + {10} &=& 9 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {2}x^2 {-1}x +{10}x {-5} $ Group the terms so that there is a common factor in each group: $ ({2}x^2 {-1}x) + ({10}x {-5}) $ Factor out the common factors: $ x(2x - 1) + 5(2x - 1) $ Notice how $(2x - 1)$ has become a common factor. Factor this out to find the answer. $(2x - 1)(x + 5)$